How is this a telescoping series?

arhzz
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Homework Statement
Finding the limit of the series
Relevant Equations
Telescoping series formula
Hello !

Consider this series;

$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$ It is said to find the limit of the series when approaches infinity.Now it is said that this is a telescopic series and that the limit is ##\frac{1}{2}## but I don't see it. I've split the an part (I don't know how to call it in english but it is the term within the sum symbol) I did that like this.

$$ \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ Now if we input k to let's say 3 we get this

$$ \frac{1}{2} + \frac{1}{3} +\frac{1}{3} + \frac{1}{5} +\frac{1}{7} + \frac{1}{7} $$. I think its fair to assume that only 1/2 should remain and the others should cancel out,that is kind of the thing by the telescoping series.But I have + across the board and the fractions won't cancel out but rather add up.I am not sure how the - is supposed to come into play here,since as I've been taught yestarteday we have to sum up the elements.The only logical solution would be that within the parantheses one element has to be negative,but I don't see how we get to that.
 
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arhzz said:
$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$ It is said to find the limit of the series when approaches infinity.Now it is said that this is a telescopic series and that the limit is ##\frac{1}{2}## but I don't see it. I've split the an part (I don't know how to call it in english but it is the term within the sum symbol) I did that like this.

$$ \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$
Are you sure about that?
 
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Why do you think

$$ \frac{1}{(2k-1)(2k+1)} = \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ ?

Plug in a value for k. Like 2.
 
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Vanadium 50 said:
Why do you think

$$ \frac{1}{(2k-1)(2k+1)} = \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ ?

Plug in a value for k. Like 2.
Okay so you are definately right,these are not the same so the algebra doesn't add up.I supposed it would have to be 1st fraction - 2nd fraction but I assumed it would be + because I don't know intuition? I saw * in the denominator and simply assumed it would be +
 
Don't use intuition. Use algebra.
 
Vanadium 50 said:
Don't use intuition. Use algebra.
Yea I guess I rushed it. It all makes sense now,thanks for pointing out the mistake.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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